
\section{Cycle Notation}

\definition{Order}
The \textit{order} of a permutation $f$ is the \textbf{smallest} positive $n$ so that $f^n(x) = x$ for all $x$. \par
If we repeatedly apply a permutation with order $n$, we will get back to where we started after $n$ steps. \par

\vspace{2mm}

For example, consider $[2134]$. This permutation has order $2$, as we clearly see below:

\begin{center}
\begin{tikzpicture}[scale=0.5]
	\node (1a) at (0, 0.5) {1};
	\node (2a) at (1, 0.5) {2};
	\node (3a) at (2, 0.5) {3};
	\node (4a) at (3, 0.5) {4};

	\node (2b) at (0, -2) {2};
	\node (1b) at (1, -2) {1};
	\node (3b) at (2, -2) {3};
	\node (4b) at (3, -2) {4};

	\node (1c) at (0, -4.5) {1};
	\node (2c) at (1, -4.5) {2};
	\node (3c) at (2, -4.5) {3};
	\node (4c) at (3, -4.5) {4};

	\line{1a}{1b}
	\line{2a}{2b}
	\line{3a}{3b}
	\line{4a}{4b}
	\line{1b}{1c}
	\line{2b}{2c}
	\line{3b}{3c}
	\line{4b}{4c}
\end{tikzpicture}
\end{center}
Swapping the first two elements of a list twice changes nothing. \par
Thus, $[2134]$ has an order of two.


\problem{}
What is the order of $[2314]$? \par
How about $[4321]$? \par
\note[Note]{You shouldn't need to draw any strings to solve this problem.}


\vfill

\problem{}
Show that all permutations (on a finite set) have a well-defined order. \par
In other words, show that there is always an integer $n$ so that $f^n(x) = x$.

\vfill

\definition{Composition}<compdef>
The \textit{composition} of two permutations $f$ and $g$ is the permutation $h(x) = f(g(x))$. \par
We'll denote this as $fg$---that is, by simply writing the permutations we're composing next to each other.

\problem{}
Show that function composition is associative. \par
That is, show that $f(gh) = (fg)h$.

\vfill

\problem{}
What is $[1324][4321]$? \par
How about $[321][213][231]$? \par


\vfill
\pagebreak


As you may have noticed, the square-bracket notation we've been using thus far is a bit unwieldy.
Permutations are verbs---but we've been referring to them using a noun (namely, their output when
applied to an ordered sequence of numbers). Our notation fails to capture the meaning of the
underlying object.

\vspace{2mm}

Think about it: is the permutation $[1234]$ different than the permutation $[12345]$? \par
Indeed, these permutations operate on different sets---but they are both the identity! \par
What should we do if we want to talk about the identity on $\{1, 2, ..., 10\}$?

\vspace{2mm}

We need something better.




\definition{Cycles}
Any permutation is composed of a number of \textit{cycles}. \par

For example, consider the permutation $[2134]$, which consists of one two-cycle: $1 \to 2 \to 1$ \par
\note[Note]{$3 \to 3$ and $4 \to 4$ are also cycles, but we'll ignore them. One-cycles aren't aren't interesting.}

\begin{center}
\begin{tikzpicture}[scale=0.5]
	\node (1) at (0, 0) {1};
	\node (2) at (1, 0) {2};
	\node (3) at (2, 0) {3};
	\node (4) at (3, 0) {4};

	\draw[line width = 0.3mm, ->, ocyan]
		(1)
		-- ($(1) + (0,-1)$)
		-- ($(2) + (0,-1)$)
		-- (2);

	\draw[line width = 0.3mm, ->, ocyan]
		(2)
		-- ($(2) + (0, 1)$)
		-- ($(1) + (0, 1)$)
		-- (1);
\end{tikzpicture}
\end{center}


The permutation $[431265]$ is a bit more interesting---it contains two cycles: \par
($1 \to 3 \to 2 \to 4 \to 1$ and $5 \to 6 \to 5$)


\begin{center}
\begin{tikzpicture}[scale=0.5]
	\node (1) at (0, 0) {1};
	\node (2) at (1, 0) {2};
	\node (3) at (2, 0) {3};
	\node (4) at (3, 0) {4};
	\node (5) at (4, 0) {5};
	\node (6) at (5, 0) {6};

	\draw[line width = 0.3mm, ->, ocyan]
		(3)
		-- ($(3) + (0,-1)$)
		-- ($(2) + (0,-1)$)
		-- (2);

	\draw[line width = 0.3mm, ->, ocyan]
		(2)
		-- ($(2) + (0,1.5)$)
		-- ($(4) + (0,1.5)$)
		-- (4);

	\draw[line width = 0.3mm, ->, ocyan]
		(4)
		-- ($(4) + (0,-1.5)$)
		-- ($(1) + (0,-1.5)$)
		-- (1);

	\draw[line width = 0.3mm, ->, ocyan]
		(1)
		-- ($(1) + (0,1)$)
		-- ($(3) + (0,1)$)
		-- (3);

	\draw[line width = 0.3mm, ->, ogreen]
		(5)
		-- ($(5) + (0,-1)$)
		-- ($(6) + (0,-1)$)
		-- (6);

	\draw[line width = 0.3mm, ->, ogreen]
		(6)
		-- ($(6) + (0,1)$)
		-- ($(5) + (0,1)$)
		-- (5);

\end{tikzpicture}
\end{center}


Another name we'll often use for two-cycles is \textit{transposition}. \par
Any permutation that swaps two adjacent elements is called a transposition. \par


\problem{}
Find all cycles in $[5342761]$.

\begin{solution}
	\begin{center}
	\begin{tikzpicture}[scale=0.5]
		\node (1) at (0, 0) {1};
		\node (2) at (1, 0) {2};
		\node (3) at (2, 0) {3};
		\node (4) at (3, 0) {4};
		\node (5) at (4, 0) {5};
		\node (6) at (5, 0) {6};
		\node (7) at (6, 0) {7};

		\draw[line width = 0.3mm, ->, ocyan]
			(1)
			-- ($(1) + (0,2)$)
			-- ($(7) + (0,2)$)
			-- (7);

		\draw[line width = 0.3mm, ->, ocyan]
			(7)
			-- ($(7) + (0,-1.5)$)
			-- ($(5) + (0,-1.5)$)
			-- (5);

		\draw[line width = 0.3mm, ->, ocyan]
			(5)
			-- ($(5) + (0,1.5)$)
			-- ($(1) + (0.5,1.5)$)
			-- ($(1) + (0.5,-1)$)
			-- ($(1) + (0,-1)$)
			-- (1);

		\draw[line width = 0.3mm, ->, ogreen]
			(2)
			-- ($(2) + (0,-1.5)$)
			-- ($(4) + (0,-1.5)$)
			-- (4);

		\draw[line width = 0.3mm, ->, ogreen]
			(4)
			-- ($(4) + (0,1)$)
			-- ($(3) + (0,1)$)
			-- (3);

		\draw[line width = 0.3mm, ->, ogreen]
			(3)
			-- ($(3) + (0,-1)$)
			-- ($(2) + (0.5,-1)$)
			-- ($(2) + (0.5,1)$)
			-- ($(2) + (0,1)$)
			-- (2);
	\end{tikzpicture}
	\end{center}
\end{solution}

\vfill


\problem{}
What permutation (on five objects) is formed by the cycles $3 \to 5 \to 3$ and $1 \to 2 \to 4 \to 1$?


\begin{solution}
	\begin{center}
	\begin{tikzpicture}[scale=0.5]
		\node (1) at (0, 0) {1};
		\node (2) at (1, 0) {2};
		\node (3) at (2, 0) {3};
		\node (4) at (3, 0) {4};
		\node (5) at (4, 0) {5};

		\draw[line width = 0.3mm, ->, ocyan]
			(3)
			-- ($(3) + (0,1)$)
			-- ($(5) + (0,1)$)
			-- (5);

		\draw[line width = 0.3mm, ->, ocyan]
			(5)
			-- ($(5) + (0,-1)$)
			-- ($(3) + (0,-1)$)
			-- (3);

		\draw[line width = 0.3mm, ->, ogreen]
			(1)
			-- ($(1) + (0,-1)$)
			-- ($(2) + (0,-1)$)
			-- (2);

		\draw[line width = 0.3mm, ->, ogreen]
			(2)
			-- ($(2) + (0,1.5)$)
			-- ($(4) + (0,1.5)$)
			-- (4);

		\draw[line width = 0.3mm, ->, ogreen]
			(4)
			-- ($(4) + (0,-1.5)$)
			-- ($(1) + (0.5,-1.5)$)
			-- ($(1) + (0.5,1)$)
			-- ($(1) + (0,1)$)
			-- (1);
	\end{tikzpicture}

	This is $[41523]$
	\end{center}
\end{solution}



\vfill
\pagebreak

\definition{Cycle Notation}
We now have a solution to our problem of notation.
Instead of referring to permutations using their output, we will refer to them using their \textit{cycles}.

\vspace{2mm}

For example, we'll write $[2134]$ as $(12)$, which denotes the cycle $1 \to 2 \to 1$:


\begin{center}
\begin{tikzpicture}[scale=0.5]
	\node (1) at (0, 0) {1};
	\node (2) at (1, 0) {2};
	\node (3) at (2, 0) {3};
	\node (4) at (3, 0) {4};

	\draw[line width = 0.3mm, ->, ocyan]
		(1)
		-- ($(1) + (0,-1)$)
		-- ($(2) + (0,-1)$)
		-- (2);

	\draw[line width = 0.3mm, ->, ocyan]
		(2)
		-- ($(2) + (0, 1)$)
		-- ($(1) + (0, 1)$)
		-- (1);
\end{tikzpicture}
\end{center}



As another example, $[431265]$ is $(1324)(56)$ in cycle notation. \par
Note that we write $[431265]$ as a \textit{composition} of two cycles: \par
applying the permutation $[431265]$ is the same as applying $(1324)$, then applying $(56)$.

\begin{center}
\begin{tikzpicture}[scale=0.5]
	\node (1) at (0, 0) {1};
	\node (2) at (1, 0) {2};
	\node (3) at (2, 0) {3};
	\node (4) at (3, 0) {4};
	\node (5) at (4, 0) {5};
	\node (6) at (5, 0) {6};

	\draw[line width = 0.3mm, ->, ocyan]
		(3)
		-- ($(3) + (0,-1)$)
		-- ($(2) + (0,-1)$)
		-- (2);

	\draw[line width = 0.3mm, ->, ocyan]
		(2)
		-- ($(2) + (0,1.5)$)
		-- ($(4) + (0,1.5)$)
		-- (4);

	\draw[line width = 0.3mm, ->, ocyan]
		(4)
		-- ($(4) + (0,-1.5)$)
		-- ($(1) + (0,-1.5)$)
		-- (1);

	\draw[line width = 0.3mm, ->, ocyan]
		(1)
		-- ($(1) + (0,1)$)
		-- ($(3) + (0,1)$)
		-- (3);

	\draw[line width = 0.3mm, ->, ogreen]
		(5)
		-- ($(5) + (0,-1)$)
		-- ($(6) + (0,-1)$)
		-- (6);

	\draw[line width = 0.3mm, ->, ogreen]
		(6)
		-- ($(6) + (0,1)$)
		-- ($(5) + (0,1)$)
		-- (5);

\end{tikzpicture}
\end{center}

Any permutation $\sigma$ may be written as a product (i.e, composition) of disjoint cycles $\sigma_1\sigma_2...\sigma_k$. \par
Make sure you believe this fact. If you don't, ask an instructor. \par
Also, the identity $f(x) = x$ is written as $()$ in cycle notation.



\problem{}
Convince yourself that disjoint cycles commute. \par
That is, that $(1324)(56) = (56)(1324) = [431265]$ since $(1324)$ and $(56)$ do not overlap. \par




\problem{}<insquare>
Write the following in square-bracket notation.
\begin{itemize}
	\item $(12)$       \tab~\tab on a set of 2 elements
	\item $(12)(435)$  \tab on a set of 5 elements
	\vspace{2mm}
	\item $(321)$      \tab~\tab on a set of 3 elements
	\item $(321)$      \tab~\tab on a set of 6 elements
	\vspace{2mm}
	\item $(1234)$      \tab on a set of 4 elements
	\item $(3412)$      \tab on a set of 4 elements
\end{itemize}
\note{
	Note that $(12)$ refers the \say{swap first two} permutation on a set of \textit{any} size. \\
	We can now use the same name for the same permutation on two different sets! \\
}

\vfill



\problem{}
Write the following in square-bracket notation.
Be careful.
\begin{itemize}
	\item $(13)(243)$ \tab on a set of 4 elements
	\item $(243)(13)$ \tab on a set of 4 elements
\end{itemize}

\vfill



\problem{}
Look at the last two permutations in \ref{insquare}, $(1234)$ and $(3412)$. \par
These are \textit{identical}---they are the same cycle written in two different ways. \par
List all other ways to write this cycle. \hint{There are two more.} \par
\note{Also, note that the last two permutations in \ref{insquare} are the same.}

\pagebreak


\problem{}
What is the inverse of $(12)$? \par
How about $(123)$? And $(4231)$? \par
\note{
	Note that again, we don't need to know how big our set is. \\
	The inverse of $(12)$ is the same in all sets.
}

\vfill


\problem{}
Say $\sigma$ is a permutation composed of disjoint cycles $\sigma_1\sigma_2...\sigma_k$. \par
Say we know the order of all $\sigma_i$. What is the order of $\sigma$?

\begin{solution}
	$\text{lcm}\Bigl(\text{ord}(\sigma_1),~ \text{ord}(\sigma_2),~ ..., ~ \text{ord}(\sigma_k)\Bigr)$
\end{solution}


\vfill

\problem{}<cycletrans>
Show that any cycle $(123...n)$ is equal to the product $(12)(23)...(n-1, n)$.

\begin{solution}
	TODO
\end{solution}

\vfill

\problem{}
Write $(7126453)$ as a product of transpositions. \par

\vfill
\pagebreak

\problem{}<simpletrans>
Show that any permutation is a product of transpositions.

\begin{solution}
	Use \ref{cycletrans}.
\end{solution}

\vfill


\problem{}
Show that any permutation is a product of transpositions of the form $(1, k)$. \par

\begin{solution}
	Use \ref{simpletrans} and rewrite each $(a, b)$ as $(1, a)(1, b)(1, a)$. \par
	Showing that $(a, b) = (1, a)(1, b)(1, a)$ is fairly easy.
\end{solution}

\vfill
\pagebreak



\problem{}
Show that any transposition $(a, b)$ is equal to the product $(a, a+1)(a+1, b)(a, a+1)$.

\begin{solution}
	This is the same as the $(1, a)(1, b)(1, a)$ case above, but we use $a + 1$
	as a \say{working slot} instead of $1$.
\end{solution}

\vfill



\problem{}
Show that any permutation is a product of adjacent transpositions. \par
(An \textit{adjacent transposition} swaps two adjacent elements, and thus looks like $(n, n+1)$)

\begin{solution}
	As before, we will use \ref{simpletrans} and rewrite the transpositions it produces in a form that fits the problem.
	We thus need to show that every transposition $(a, b)$ is a product of adjacent transpositions.

	\vspace{8mm}

	In the proof below, assume that $a < b$ and perform induction on $b - a$. \par

	\textbf{Base Case:}\par
	If $b - a = 1$, we clearly see that $(a, b)$ is a product of adjacent. \par
	In fact, it \textit{is} an adjacent transposition.

	\vspace{4mm}

	\textbf{Induction:}\par
	Now, say $b - a = n + 1$. \par
	Assume that all $(a, b)$ where $b - a \leq n$ are products of adjacent transpositions.\par
	Note that $(a, b) = (a, a+1)(a+1, b)(a, a+1)$.

	\vspace{2mm}

	$(a, a+1)$ is an adjacent transposition, and $b - (a+1) = n$. \par
	Thus, $(a, b)$ is a product of adjacent transpositions.
\end{solution}


\vfill
\pagebreak